TITLE: Let $\left\{\Delta_1,\Delta_2,.....,\Delta_n\right\}$ be the set of all determinants of order 3 that can be made with the distinct real numbers
QUESTION [12 upvotes]: Let $\left\{\Delta_1,\Delta_2,.....,\Delta_n\right\}$ be the set of all determinants of order 3 that can be made with the distinct real numbers from the set $S=\left\{1,2,3,4,5,6,7,,8,9\right\}$.Then prove that $\sum_{i=1}^{n}\Delta_i=0$

I know that the total number of determinants that can be formed by using distinct real numbers from the given set is $9!$.But i dont know how to prove that their sum is zero.Please help me.Thanks

REPLY [6 votes]: We have $det(A) = a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} -a_{11}a_{23}a_{32} - a_{13}a_{22}a_{31} - a_{12}a_{21}a_{33}$. There are $9!$ such sums. So each term in this sum will be the same (eg. $\Sigma a_{11}a_{22}a_{33} = \Sigma a_{12}a_{23}a_{31}$) , and there are exactly equal number of positive and negative terms, so the result is zero.